Question

Two vibrating strings of the same material but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one length $L$ with frequency ${\upsilon _1}$ and the other with frequency ${\upsilon _2}$. Find the ratio $\dfrac{{{\upsilon _1}}}{{{\upsilon _2}}}$?

Answer 1

Hint: Since, two strings of the same material vibrates. Thus, the density of both the strings is the same. Hence, find the linear density or the mass per unit length of both the strings. The vibration of string produces the transverse wave, thus finding the velocity of the transverse wave for both the strings. Since, they vibrate at a fundamental node, thus the wavelength is twice the length of the wire. Finally use the relation between the velocity, wavelength and frequency of the wave to obtain the ration of their frequencies.

Formula used:
Mass per unit length is given by,
$\mu = \dfrac{M}{L}$
Where, $\mu$ is the mass per unit length and $M$ is the mass of string

Mass of the string is given by,
$M = V \times \rho$
Where, $V$ is the volume of string and $\rho$ is the density.

Velocity of transverse wave is given by,
$v = \sqrt {\dfrac{T}{\mu }}$
Where, $v$ is velocity, $T$ is the tension on string and $\mu$ is the mass per unit length.

Relation between the velocity, wavelength and frequency,
$\upsilon = \dfrac{v}{\lambda }$
Where, $\upsilon$ is the frequency, $v$ is the velocity and $\lambda$ is the wavelength.

Complete step by step solution:
Mass of the first string is,
${M_1} = {V_1} \times \rho$
Since, $V = \pi {r^2}L$
${M_1} = \pi {\left( {2r} \right)^2}L\rho \\\ {M_1} = 4\pi {r^2}L\rho \\\$

Mass of the second string is,
${M_2} = {V_2} \times \rho$
Since, $V = \pi {r^2}L$
${M_2} = \pi {\left( r \right)^2}2L\rho \\\ {M_2} = 2\pi {r^2}L\rho \\\$

Mass per unit length for first string is given by
${\mu _1} = \dfrac{{{M_1}}}{L}$
Substitute the value of ${M_1}$ in above equation,
${\mu _1} = \dfrac{{4\pi {r^2}L\rho }}{L} \\\ {\mu _1} = 4\pi {r^2}\rho \\\$

Mass per unit length for second string is given by
${\mu _2} = \dfrac{{{M_2}}}{L}$
Substitute the value of ${M_2}$ in above equation,
${\mu _2} = \dfrac{{2\pi {r^2}L\rho }}{{2L}} \\\ {\mu _2} = \pi {r^2}\rho \\\$

Since, the tension is same ${T_1} = {T_2} = T$
Velocity of transverse wave of first string is given by,
${v_1} = \sqrt {\dfrac{T}{{{\mu _1}}}}$
Substitute the value of ${\mu _1}$ in above equation,
${v_1} = \sqrt {\dfrac{T}{{4\pi {r^2}\rho }}}$
Velocity of transverse wave of second string is given by,
${v_2} = \sqrt {\dfrac{T}{{{\mu _2}}}}$
Substitute the value of ${\mu _2}$ in above equation,
${v_2} = \sqrt {\dfrac{T}{{\pi {r^2}\rho }}}$
In fundamental mode, the wavelength of wire is equal to twice the length of the wire,
Hence,
Wavelength of first wire, ${\lambda _1} = 2L$
Wavelength of second wire, ${\lambda _2} = 4L$

Relation between the velocity, wavelength and frequency,
$\upsilon = \dfrac{v}{\lambda }$
For the first string,
${\upsilon _1} = \dfrac{{{v_1}}}{{{\lambda _1}}}$
Substitute the values in above equation,
${\upsilon _1} = \dfrac{{{v_1}}}{{2L}} \\\ {\upsilon _1} = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{{4\pi {r^2}\rho }}} \\\ {\upsilon _1} = \dfrac{1}{{4L}}\sqrt {\dfrac{T}{{\pi {r^2}\rho }}} \\\$

For the second string,
${\upsilon _2} = \dfrac{{{v_2}}}{{{\lambda _2}}}$
Substitute the values in above equation,
${\upsilon _2} = \dfrac{{{v_2}}}{{4L}} \\\ {\upsilon _2} = \dfrac{1}{{4L}}\sqrt {\dfrac{T}{{\pi {r^2}\rho }}} \\\ \\\$
Hence, the ratio between ${\upsilon _1}$ and ${\upsilon _2}$,
$\dfrac{{{\upsilon _1}}}{{{\upsilon _2}}} = \dfrac{{\dfrac{1}{{4L}}\sqrt {\dfrac{T}{{\pi {r^2}\rho }}} }}{{\dfrac{1}{{4L}}\sqrt {\dfrac{T}{{\pi {r^2}\rho }}} }} \\\ \dfrac{{{\upsilon _1}}}{{{\upsilon _2}}} = 1 \\\$

Thus, the option (D) is correct.

Note: In the given question, both strings have the same density and have different radii and length. Thus using the various formulas of linear mass, the velocity of string and the relation between velocity, wavelength and frequency, the final ratio between the frequencies is derived.